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Beecrowd 1095 solution || How to Solve the Sequence IJ-1 Problem beecrowd 1095 solution || Free Code Center

Beecrowd 1095 solution

Sequence IJ 1

Introduction

In this article, we will be discussing the URI Bee Crowd 1095 Sequence IJ 1 solution in C/C++/Python. We will provide a detailed explanation of the problem, walk you through the solution step by step, and provide code snippets in C/C++/Python to help you implement the solution. By the end of this article, you will have a better understanding of the problem and be able to solve it with ease.

Problem Description

The URI Bee Crowd 1095 Sequence IJ 1 problem requires us to find the sum of all even numbers between two integers I and J, inclusive. The values of I and J will be given as input, and we need to output the sum of all even numbers between I and J. If I or J is an odd number, we need to consider the next even number as the starting or ending point.

Implementation in C

#include<stdio.h>

int main()
{
    int i, j, sum = 0;
    scanf("%d%d", &i, &j);
    if(i%2 == 1) i++;
    for(; i <= j; i+=2)
    {
        sum += i;
    }
    printf("%d\n", sum);
    return 0;
}

Implementation in Cpp

#include <iostream>

using namespace std;

int main() {

   // cout<<"I=1"<<" "<<"j=60"<<endl;
    for(int i=1,j=60;i<15,j>=0;i+=3,j-=5){


       cout<<"I="<<i <<" "<<"J="<<j<<endl;

    }

    return 0;
}

Beecrowd 1095 solution in Python

Implementation in Python

i, j = map(int, input().split())
if i%2 == 1: i += 1
print(sum(range(i, j+1, 2)))

Conclusion

In conclusion, we have provided a detailed explanation of the URI Bee Crowd 1095 Sequence IJ 1 problem and provided solutions in C/C++/Python. By implementing the solutions provided in this article, you will be able to solve the problem with ease. We hope this article has been helpful, and if you have any further questions or comments, please feel free to leave them below.

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Next problems: Beecrowd 1178 Array Fill III Solution in C, C++ ,Beecrowd 1049 Animal Solution

Beecrowd 1178 Array Fill III Solution in C, C++|| Beecrowd 1178 || Free Code Center

Problem link

Beecrowd 1178 Array Fill III Solution in C, C++

Read a number X. Put this X at the first position of an array [100]. In each subsequent position (1 up to 99) put half of the number inserted at the previous position, according to the example below. Print all the vector N.

Input

The input contains a double-precision number with four decimal places.

Output

For each position of the array N print “N[i] = Y”, where i is the array position and Y is the number stored in that position. Each number of N[…] must be printed with 4 digits after the decimal point.

Input SampleOutput Sample
200.0000N[0] = 200.0000
N[1] = 100.0000
N[2] = 50.0000
N[3] = 25.0000
N[4] = 12.5000

Beecrowd 1178 Array Fill III Solution in C++

#include<iostream>
#include <iomanip>

using namespace std;
int main()
{
   double a;
   cin>>a;
   for(int i=0;i<100;i++){

               cout<<"N["<<i<<"] = "<<setprecision(4)<<fixed<<a<<endl;

         a=a/2.0;
             //  cout<<"N["<<i<<"]"<<setprecision(4)<<fixed<<a<<endl;


   }
   return 0;
}

Next problems:URI Bee Crowd 1095: How to Solve the Sequence IJ-1 Problem

Uri 1113 Solution || Beecrowd 1113 Ascending and Descending Solution || Beecrowd 1113 || Free Code Center

Uri 1113 solution || Beecrowd1113 Ascending and Descending Solution

Ascending and Descending

Introduction

URI Online Judge is a popular platform among developers for practicing coding skills. One of the problems on the platform is “Bee Crowd – 1113 Ascending and Descending,” and in this article, we will provide a comprehensive solution to the problem in C++, C, and Python.

Understanding the Problem

The “Bee Crowd – 1113 Ascending and Descending” problem requires us to sort an array of integers in ascending and descending orders. We need to implement two separate sorting algorithms for both ascending and descending orders.

Solution in C

#include<stdio.h>
int main()
{
    int x, y;
    while(1)
    {
        scanf("%d%d", &x, &y);
        if(x==y)
        {
            break;
        }
        else if(x>y)
        {
            printf("Decrescente\n");//uri 1113 solution
        }
        else
        {
            printf("Crescente\n");
        }
    }
    return 0;
}

Solution in C++

#include<iostream>
using namespace std;
int main()
{
    int x,y;
    while(1)
    {
        cin>>x>>y;
        if(x==y)
        {
            break;
        }

     else if(x>y)
    {
        cout<<"Decrescente"<<endl;//Uri 1113 solution
    }
    else
    {
        cout<<"Crescente"<<endl;
    }
    }
    return 0;
}

Previous problem: Beecrowd 1161 Factorial Sum Solution

Beecrowd 1161 Factorial Sum Solution

Beecrowd 1161 Factorial Sum Solution in C, C++, Java

Welcome to our comprehensive guide on URI Factorial Sum Solutions in C, C++, and Java. In this guide, we will provide you with an in-depth understanding of the factorial sum problem and its solutions in these three programming languages. We will also compare and contrast the solutions to help you choose the best option for your specific needs.

What is URI Factorial Sum Problem?

The URI Factorial Sum Problem is a mathematical problem that requires us to calculate the sum of factorials of a given number. The problem is stated as follows:

Given a number n, find the sum of factorials of all the numbers from 1 to n.

For example, if n is 3, then the sum of factorials would be 1! + 2! + 3! = 1 + 2 + 6 = 9.

Bee1161 Solutions in C:

To solve the URI Factorial Sum problem in C, we can use loops to iterate through each number from 1 to n and calculate the factorial sum. Here is the code snippet to solve the problem in C:

#include<stdio.h>

int main()
{
    int n, i, fact = 1, sum = 0;
    
    printf("Enter the value of n: ");
    scanf("%d", &n);
    
    for(i=1;i<=n;i++)
    {
        fact = fact * i;
        sum = sum + fact;
    }
    
    printf("Sum of factorials of 1 to %d is %d", n, sum);
    
    return 0;
}

Bee1161 Solutions in C++:

In C++, we can use a similar approach as C to solve the URI Factorial Sum problem. However, C++ provides us with some additional features such as the iostream library and the using namespace std; statement. Here is the code snippet to solve the problem in C++:

#include <iostream>
using namespace std;

int main() {

   long long int m,n,f=1,f1=1;
   while(cin>>m>>n){
   for(int i=1;i<=m;i++)
   {
       f=f*i;
   }
   for(int j=1;j<=n;j++)
   {
       f1=f1*j;
   }

   long long int sum=f+f1;
   cout<<sum<<endl;
   f=f1=1;

   }
    return 0;
    cout<<endl;
}

Bee1161 Solutions in Java:

In Java, we can use the BigInteger class to handle large factorials. Here is the code snippet to solve the URI Factorial Sum problem in Java:

import java.math.BigInteger;
import java.util.Scanner;

public class FactorialSum {

    public static void main(String[] args) {
        
        Scanner input = new Scanner(System.in);
        
        int n = input.nextInt();
        BigInteger fact = BigInteger.ONE, sum = BigInteger.ZERO;
        
        for(int i=1;i<=n;i++)
        {
            fact = fact.multiply(BigInteger.valueOf(i));
            sum = sum.add(fact);
        }
        
        System.out.println("Sum of factorials of 1 to "+n+" is "+sum);
    }

}

Comparison of Solutions:

All three solutions provide an efficient and effective way to solve the URI Factorial Sum problem. However, Java’s BigInteger the class provides us with the ability to handle very large factorials that are not possible with C and C++. C++ is slightly faster than C, but the difference is negligible for small inputs.

Conclusion:

In this guide, we provided you with a comprehensive understanding of the URI Factorial Sum problem and its solutions in C, C++, and Java. We hope this guide helps you choose the best solution for your specific needs.

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Beecrowd 1177 Array Fill II solution

Beecrowd 1177 Array Fill II solution

Introduction:

Programming problems like URI 1177 Array Fill II are frequently used to evaluate the coding skills of computer science students. Students must create a program that inserts a particular value into an array and outputs the array’s values in a particular format. We will provide a thorough explanation of how to solve URI 1177 Array Fill II in this article, including with sample code, directions, and advice to help you finish the task.

Section 1:

Recognizing the Issue Understanding the problem statement in its whole is crucial for solving Beecrowd 1177 Array Fill II solution. You must first add a certain value X to an array of size N before printing the array in a specific format.The message “N[0] = X” should appear on the first line of the output, followed by “N[1] = X+1” on the second line, and so on, printing each element of the array one by one. A for loop can be used to run through the array and print each element while adhering to the desired format.

Section 2:

Guide to the Solution in Steps Use these procedures to resolve URI 1177 Array Fill II:


N’s value should be read from the input.

Make an array of N elements.

X’s value should be read from the input.//uri 1177 solution

Using a for loop to cycle through each element of the array, assign the value X to each element of the array.

Using another for loop to run through each element and display it in the necessary format, print every element of the array.

Solution in C

#include <stdio.h>

int main() {

int a[1000],i,j,n;
scanf("%d",&n);
for(i=0,j=0;i<1000;i++){

        printf("N[%d] = %d\n",i,j);//uri 1177 solution
        j++;
       if(n==j) j=0;
        }

    return 0;
}

Conclusion:

Even though URI 1177 Array Fill II is a difficult programming issue, you can overcome it by using this in-depth solution manual. You can improve your comprehension of the issue and learn how to develop efficient code to address it by carefully following the step-by-step instructions and using the sample code that is offered. We sincerely hope that this guide has been useful to you and wish you luck in completing URI 1177 Array Fill II.

Previous problem: Beecrowd 1176 Fibonacci Array Solution

Beecrowd 1176 Fibonacci Array Solution

Beecrowd 1176 Fibonacci Array Question

Beecrowd 1176 Fibonacci Array Solution.Understanding Fibonacci Sequences: Before we dive into solving the problem, it is important to understand what Fibonacci sequences are. Fibonacci sequences are a series of numbers in which each number is the sum of the two preceding numbers. The sequence starts with 0 and 1, and the next number in the sequence is the sum of the two preceding numbers. For example, the first 10 numbers in the Fibonacci sequence are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34.

Solving the URI 1176 Fibonacci Array Problem: To solve the URI 1176 Fibonacci Array problem, we need to first generate a Fibonacci sequence up to a given number. Once we have the sequence, we can calculate the sum of all even numbers in the sequence.

#include <stdio.h>

int main() {

   int i,j,a,d;
   long long int arr[61];
   arr[0]=0;
   arr[1]=1;
   for(i=2;i<61;i++)
       arr[i]=arr[i-1]+arr[i-2];
       scanf("%d",&a);
       for(j=0;j<a;j++){
           scanf("%d",&d);
           printf("Fib(%d) = %lld\n",d,arr[d]);
       }


    return 0;
}
#include <iostream>
using namespace std;

int main() {
    int i, j, a, d;
    long long int arr[61];
    arr[0] = 0;
    arr[1] = 1;

    for (i = 2; i < 61; i++)
        arr[i] = arr[i - 1] + arr[i - 2];

    cin >> a;

    for (j = 0; j < a; j++) {
        cin >> d;
        cout << "Fib(" << d << ") = " << arr[d] << endl;
    }

    return 0;
}

Previous problem: Even Square Problems Solution/Beecrowd 1073

Beecrowd 1073 Even Square Problems Solution

Beecrowd 1073 Even Square Problems Solution in C++, C, and Python

Even Square

Read an integer N. Print the square of each one of the even values from 1 to including if it is the case.

Input

The input contains an integer (5 < < 2000).

Output

Print the square of each one of the even values from 1 to N, as the given example.

Be carefull! Some language automaticly print 1e+006 instead 1000000. Please configure your program to print the correct format setting the output precision.

Input SampleOutput Sample
62^2 = 4
4^2 = 16
6^2 = 36
 

First, let’s take a closer look at the problem itself. The URI BeeCrowd Bee 1073 Even Square problem asks you to find all the even numbers between 1 and a given number, and then print each number and its square on a separate line. This is a fairly simple problem, but it does require some basic coding skills and knowledge of loops and conditionals.

To solve this problem in C, C++, or Python, you’ll need to use a loop to iterate over all the even numbers between 1 and the given number. Then, for each even number, you’ll need to calculate its square and print the result to the console. We’ll provide detailed code examples in each language below.

C Solution

#include <stdio.h>

int main() {

    int i,n,s=1;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        if(i%2==0){
           s=i*i;
           printf("%d^2 = %d\n",i,s);
        }


    }

    return 0;
}

In this C solution, we use a for loop to iterate over all the even numbers between 2 and the given number. We start at 2 instead of 1 because we know that 1 is not an even number. We increment the loop variable i by 2 at each iteration, which ensures that we only visit even numbers. Inside the loop, we print out the square of the current number using the %d format specifier for integers.

C++ Solution

#include <iostream>

using namespace std;

int main() {
    int n;
    cin >> n;
    for(int i = 2; i <= n; i += 2) {
        cout << i << "^2 = " << i*i << endl;
    }
    return 0;
}

In this C++ solution, we use a similar approach to the C solution, but we use cout and cin instead of printf and scanf. We also use the namespace keyword to avoid having to prefix cout and cin with std:: every time.

Python Solution

n = int(input())
for i in range(2, n+1, 2):
    print(f'{i}^2 = {i*i}')

In this Python solution, we use a for loop and the range function to iterate over all the even numbers between 2 and the given number. We use an f-string to print out the square of the current number.

Conclusion

In this guide, we’ve provided detailed solutions to the URI BeeCrowd Bee 1073 Even Square problem in C, C++, and Python. We’ve also discussed some basic coding concepts and provided code examples in each language. We believe that this guide is the best resource available for solving this particular problem, and we hope that it helps you in your coding journey. Thank you for choosing Codeshikhi, and happy coding!

Beecrowd All solution

Beecrowd 1071 Sum of Consecutive Odd Numbers I Solution || Beecrowd 1071

Beecrowd 1071 Sum of Consecutive Odd Numbers I Solution

Introduction

The URI 1071 problem requires finding the sum of all the odd numbers between two input values. This is a common problem in programming, and it is often used to test a programmer’s skills in loop constructs, conditional statements, and basic arithmetic operations. In this article, we will discuss the problem in detail, provide step-by-step solutions in C, C++, C++, Python, and Java, and explain the underlying logic and concepts.

Understanding the Problem

The URI 1071 problem states that we need to find the sum of all odd numbers between two input values. To solve this problem, we need to use a loop construct that iterates through all numbers between the two input values and checks whether they are odd or not. If a number is odd, we need to add it to the running sum.

Solution

We can solve the URI 1071 problem in several programming languages, including C, C++, C++, Python, and Java. Here is a step-by-step guide on how to solve the problem in each language.

C Solution

#include<stdio.h>
int main(){
    int a,b,sum=0;
    scanf("%d %d",&a,&b);
    if(a>b){
        int temp=a;
        a=b;
        b=temp;
    }
    for(int i=a+1;i<b;i++){
        if(i%2!=0) sum+=i;
    }
    printf("%d\n",sum);
    return 0;
}

C++ Solution

#include<bits/stdc++.h>
using namespace std;
int main(){
    int a,b,sum=0;
    cin>>a>>b;
    if(a>b) swap(a,b);
    for(int i=a+1;i<b;i++){
        if(i%2!=0) sum+=i;
    }
    cout<<sum<<"\n";
    return 0;
}

Python Solution

a=int(input())
b=int(input())
if a>b:
    a,b=b,a
sum=0
for i in range(a+1,b):
    if i%2!=0: sum+=i
print(sum)

Java Solution

import java.util.*;
class Main{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int a,b,sum=0;
        a=sc.nextInt();
        b=sc.nextInt();
        if(a>b){
            int temp=a;
            a=b;
            b=temp;
        }
        for(int i=a+1;i<b;i++){
            if(i%2!=0) sum+=i;
        }
        System.out.println(sum);
    }
}

Explanation

The solutions provided above use a loop construct to iterate through all numbers between the two input values, a and b. We first check which value is larger and then swap them if needed, so that a is always the smaller of the two values. We then use a for loop to iterate through all numbers between a+1 and b-1, and check whether each number is odd or not. If the number is odd, we add it to the running sum. Finally, we print the sum of all odd numbers between a and b.

Conclusion

In conclusion, the URI 1071 problem requires finding the sum of all odd numbers between

BEECROWD Problems Solution

Beecrowd-1070 Six Odd Numbers Solution

Beecrowd-1070 Six Odd Numbers Solution

Beecrowd 1070

Six Odd Number

Read an integer value and print the 6 consecutive odd numbers from X, a value per line, including if it is the case.

Input

The input will be a positive integer value.

Output

The output will be a sequence of six odd numbers.

Input SampleOutput Sample
89
11
13
15
17
19

Beecrowd 1070 Six Odd Numbers Solution in C

#include<stdio.h>
int main()
{
    int x,i;
    scanf("%d",&x);
     if(x%2==0)x++;
    for(i=0;i<6;i++){

        printf("%d\n",x);
        x=x+2;

    }
    return 0;
}

Solution in C++

#include<iostream>
using namespaces std;
int main()
{
int x;
cin>>x;
if(x%2==0)x++;
for(int i=0;i<6;i++)
{ cout<<x<<endl;
x=x+2;
}
return 0;
}

Beecrowd All solution

Beecrowd 1175 – Array change I Solution

Beecrowd 1175 – Array change I Solution

Array change I

URI 1175 Array change I

Problem:

Write a program that reads an array N [20]. After, change the first element by the last, the second element by the penultimate, etc., until changing the 10th to the 11th. Print the modified array.

Input:

The input contains 20 integer numbers, positive or negative.

Output:

For each position of the array N print “N[i] = Y”, where i is the array position and Y is the number stored in that position after the modifications.

Beecrowd 1175 - Array change I

Beecrowd 1175 – Array change I solution in C

#include<stdio.h>
int main() {
 int N[20],i,temp=0;
for(i=19;i>=0;i--){
    scanf("%d",&N[i]);
}



for(i=0;i<20;i++){

    printf("N[%d] = %d\n",i,N[i]);
}

    return 0;
}

Explanation:

In this program, we first declare an integer array N of size 20 and also declare three integer variables i, j, and temp. We use a for loop to take input from the user for the array N.

Next, we use another for loop to swap the values of the array elements. In each iteration, we take the ith element and the jth element and swap their values. We use a temporary variable temp to store the value of the ith element before swapping. We start swapping from the beginning and end of the array and move towards the middle until we reach the 10th and 11th elements.

Finally, we use another for loop to print the modified array. We print each element of the array using the printf function.

This solution uses basic array manipulation and should be easy to understand.

BEECROWD Problems Solution